Question

[Nguyễn Anh Tú]

Giúp bài 5 vs 6 á

Answer 1

[Nguyễn Việt Lâm]

5.

Gọi M là trung điểm BC

\(\Rightarrow\left\{{}\begin{matrix}x_M=\dfrac{x_B+x_C}{2}=\dfrac{2+\left(-1\right)}{2}=\dfrac{1}{2}\\y_M=\dfrac{y_B+y_C}{2}=\dfrac{1+\left(-2\right)}{2}=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow M\left(\dfrac{1}{2};-\dfrac{1}{2}\right)\)

Gọi G là trọng tâm tam giác

\(\Rightarrow\left\{{}\begin{matrix}x_G=\dfrac{x_A+x_B+x_C}{3}=\dfrac{4}{3}\\y_G=\dfrac{y_A+y_B+y_C}{3}=1\end{matrix}\right.\) \(\Rightarrow G\left(\dfrac{4}{3};1\right)\)

b.

Gọi \(D\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=\left(-1;-3\right)\\\overrightarrow{DC}=\left(-1-x;-2-y\right)\end{matrix}\right.\)

ABCD là hbh khi

\(\overrightarrow{AB}=\overrightarrow{DC}\Rightarrow\left\{{}\begin{matrix}-1-x=-1\\-2-y=-3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\)

\(\Rightarrow D\left(0;1\right)\)

Answer 2

[Nguyễn Việt Lâm]

6.

\(\left\{{}\begin{matrix}x_G=\dfrac{x_A+x_B+x_C}{3}=\dfrac{1}{3}\\y_G=\dfrac{y_A+y_B+y_C}{3}=0\end{matrix}\right.\)

\(\Rightarrow G\left(\dfrac{1}{3};0\right)\)

b.

Gọi \(D\left(x;y\right)\) \(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{BG}=\left(-\dfrac{26}{3};10\right)\\\overrightarrow{DC}=\left(-5-x;4-y\right)\end{matrix}\right.\)

BGCD là hình bình hành khi:

\(\overrightarrow{BG}=\overrightarrow{DC}\Rightarrow\left\{{}\begin{matrix}-5-x=-\dfrac{26}{3}\\4-y=10\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{11}{3}\\y=-6\end{matrix}\right.\)

\(\Rightarrow D\left(\dfrac{11}{3}=-6\right)\)